Question

Activity_Mass_Balance_Recycle

Sea water must be desalinated by reverse osmosis. For a feed rate of 1000 lb/h sea water containing 3.1% salt (by weight) is desalinated water with only 500 ppm salt and the brine tailings containing 5.25% salt. Part of the brine tailings is recycled and the inlet stream in the reverse osmosis cell contains 4.0% salt. Determine:

a) A brine removal taxon (ṁ3);
b) The production rate of desalinated water (ṁ2);
c) The rate of brine that is recycled (ṁR).

Answers:

ṁ1 = 1720 lb / h

ṁ2 = 409.37 lb / h

ṁ3 = 590.63 lb / h

ṁR = 720 lb / h

Answer 1

m₁ = 1720

m₂ = 413.46

m₃ = 586.54

mr = 720

Step-by-step explanation:

Draw a diagram. Sea water (1000 lb/h, 3.1% salt) is mixed with the recycled brine (mr, 5.25% salt) to create the inlet stream (m₁, 4.0% salt). This inlet stream is fed into the osmosis cell. The osmosis cell produces desalinated water (m₂, 500 ppm) and brine tailings. Part of the tailings is recycled (mr, 5.25% salt), and the rest is removed (m₃, 5.25%).

The mass of water is conserved, so:

1000 + mr = m₁

m₁ = mr + m₂ + m₃

The mass of salt is conserved, so:

0.031 (1000) + 0.0525 mr = 0.040 m₁

0.040 m₁ = 0.0525 mr + 0.0005 m₂ + 0.0525 m₃

Four unknown variables, and four equations. We can use the first and third to find mr and m₁.

0.031 (1000) + 0.0525 mr = 0.040 (1000 + mr)

31 + 0.0525 mr = 40 + 0.040 mr

0.0125 mr = 9

mr = 720

m₁ = 1720

Now we can use these values and the second and fourth equations to solve for the remaining variables.

1720 = 720 + m₂ + m₃

0.040 (1720) = 0.0525 (720) + 0.0005 m₂ + 0.0525 m₃

1000 = m₂ + m₃

68.8 = 37.8 + 0.0005 m₂ + 0.0525 m₃

68.8 = 37.8 + 0.0005 m₂ + 0.0525 (1000 − m₂)

68.8 = 37.8 + 0.0005 m₂ + 52.5 − 0.0525 m₂

0.052 m₂ = 21.5

m₂ = 413.46

m₃ = 586.54